Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& v) {
  int ss = ((int)(v).size());
  for (int(i) = 0; (i) < (ss); ++(i)) out << v[i] << " ";
  return out;
}
template <typename T, typename TT>
ostream& operator<<(ostream& out, const pair<T, TT>& p) {
  out << p.first << " " << p.second;
  return out;
}
template <typename T, typename TT>
istream& operator>>(istream& in, pair<T, TT>& s) {
  in >> s.first >> s.second;
  return in;
}
template <typename T>
bitset<sizeof(T) * 8> BIN(T a) {
  return bitset<sizeof(T) * 8>(a);
}
vector<int> pq[20][100003];
int n, m, q;
int par[100003][20];
int dep[100003];
vector<int> people[100003];
vector<int> adj[100003];
void dfs(int on, int pare, int depth) {
  par[on][0] = pare;
  dep[on] = depth;
  for (auto it : adj[on]) {
    if (it != pare) {
      dfs(it, on, depth + 1);
    }
  }
}
vector<int> merge(const vector<int>& a, const vector<int>& b) {
  int p1 = 0, p2 = 0;
  int x = ((int)(a).size());
  int y = ((int)(b).size());
  vector<int> v;
  for (int i = 0; i < 10; i++) {
    if (p1 < x) {
      if (p2 < y) {
        if (a[p1] < b[p2]) {
          v.push_back(a[p1]);
          p1++;
        } else {
          v.push_back(b[p2]);
          p2++;
        }
      } else {
        v.push_back(a[p1]);
        p1++;
      }
    } else {
      if (p2 < y) {
        v.push_back(b[p2]);
        p2++;
      } else {
        break;
      }
    }
  }
  return v;
}
void build() {
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      par[j][i] = par[par[j][i - 1]][i - 1];
      pq[i][j] = merge(pq[i - 1][j], pq[i - 1][par[j][i - 1]]);
    }
  }
}
vector<int> get(int a, int b) {
  if (dep[a] < dep[b]) {
    swap(a, b);
  }
  vector<int> q;
  int difff = dep[a] - dep[b];
  for (int i = 0; i < 21; i++) {
    if ((difff >> i) & 1) {
      q = merge(q, pq[i][a]);
      a = par[a][i];
    }
  }
  if (a == b) {
    q = merge(q, pq[0][b]);
    return q;
  }
  for (int i = 19; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      q = merge(q, merge(pq[i][a], pq[i][b]));
      a = par[a][i];
      b = par[b][i];
    }
  }
  return merge(q, merge(merge(pq[0][a], pq[0][b]), pq[0][par[a][0]]));
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    int c;
    scanf("%d", &c);
    pq[0][c].push_back(i + 1);
  }
  for (int(i) = 1; (i) <= (n); ++(i)) {
    if (!pq[0][i].empty()) {
      sort(pq[0][i].begin(), pq[0][i].end());
    }
  }
  dfs(1, 0, 0);
  build();
  while (q--) {
    int v, u, a;
    scanf("%d%d%d", &v, &u, &a);
    auto ans = get(v, u);
    int val = min(a, ((int)(ans).size()));
    printf("%d", val);
    for (int(i) = 0; (i) < (val); ++(i)) {
      printf(" %d", ans[i]);
    }
    printf("\n");
  }
  return 0;
}